# Office floor assignment

A real estate developer has built a new office building of six floors, and numbered the floors 1 through 6 from bottom to top. Each of the six companies — F, G, I, J, K, and M — must be assigned an entire floor for office space. The floors must be assigned according to the following conditions:

• F must be on a floor lower than G
• I must be either on the floor immediately above M or on the floor immediately below M
• J can be neither on the floor immediately above M not the floor immediately below M
• K must be on floor 4

Q7: Which of the following is an acceptable assignment of companies to floors 1 through 6?

G, I, M, K, F, J

J, F, G, K, I, M

J, M, I, F, G, K

K, F, J, G, M, I

Q8: If M is on floor 2, any one of the following could be true EXCEPT:

F is on floor 3

I is on floor 1

K is on floor 5

J is on floor 6

Q9: If J is on floor 3, which of the following pairs of companies cannot be on consecutive floors?

M and J

F and K

G and J

K and M

Q10: If F is on floor 5, which of the following cannot be true?

G is on floor 4

K is on floor 4

M is on floor 2

M is on floor 3

• Can you please edit by changing company 'I' to a different alphabet for better readability:-) – Mea Culpa Nay Sep 28 '17 at 5:17
• What about Q 1-6? Where did you get this questions from? here at Puzzling we like to attribute our sources. – micsthepick Sep 28 '17 at 5:29
• this 'puzzle' is ridiculously easy. – JMP Sep 28 '17 at 7:17
• Did this come from an LSAT? – Acccumulation Sep 28 '17 at 17:45

For No. 7:

Instructions say that F must be lower than G. and K must be at 4. So, only option 2 satisfies the required condition.

For no. 8:

Again K must be at 4. So, Option 3 is always wrong (Hence, the correct answer).

For no. 9:

Option 1 is again wrong (and thus the correct option) because it violates the condition that J and M can't be in consecutive floors.

For no. 10:

Option 1 can't be true (and again, the correct option) because G has to be higher than F. And if, F is at 5, G has to be at 6. Alternative method is that K is always at 4. Hence, G can never be at 4. That also leads to the same answer.